62.Unique-Paths

62. Unique Paths

题目地址

https://leetcode.com/problems/unique-paths/

https://www.jiuzhang.com/solutions/unique-paths

题目描述

A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below).

The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in the diagram below).

How many possible unique paths are there?

Example 1:
Input: m = 3, n = 2
Output: 3
Explanation:
From the top-left corner, there are a total of 3 ways to reach the bottom-right corner:
1. Right -> Right -> Down
2. Right -> Down -> Right
3. Down -> Right -> Right

Example 2:
Input: m = 7, n = 3
Output: 28

Constraints:
1 <= m, n <= 100
It's guaranteed that the answer will be less than or equal to 2 * 10 ^ 9.

代码

Approach #1 Recursion

class Solution {
  public int uniquePaths(int m, int n) {
    if (m == 1 || n == 1)        return 1;

    return uniquePaths(m - 1, n) + uniquePaths(m , n - 1);
  }
}

Approach #2 Dynamic Programming

class Solution {
    public int uniquePaths(int m, int n) {
        int[][] dp = new int[m][n];
        for (int[] arr: dp) {
            Arrays.fill(arr, 1);
        }

        for (int row = 1; row < m; row ++) {
            for (int col = 1; col < n; col++) {
                dp[row][col] = dp[row - 1][col] + dp[row][col - 1];
            }
        }

        return dp[m - 1][n - 1];
    }
}

#2

public class Solution {
    public int uniquePaths(int m, int n) {
        if (m == 0 || n == 0) return 1;

        int[][] sum = new int[m][n];
        for (int i = 0; i < m; i++) {
            sum[i][0] = 1;
        }
        for (int i = 0; i < n; i++) {
            sum[0][i] = 1;
        }

        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                sum[i][j] = sum[i - 1][j] + sum[i][j - 1];
            }
        }
        return sum[m - 1][n - 1];
    }
}

#3

public class Solution {
    public int uniquePaths(int m, int n) {
        if (m == 0 || n == 0) return 1;

        int[][] sum = new int[m][n];
        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                if (i == 0 || j == 0) {
                    sum[i][j] = 1;
                } else {
                    sum[i][j] = sum[i - 1][j] + sum[i][j - 1];
                }
            }
        }
        return sum[m - 1][n - 1];
    }
}